Question

Let $\displaystyle\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$, where $a, b, c \in Z$ and $e=\displaystyle\sum_{n=0}^{\infty} \frac{1}{n !}$ Then $a^2-b+c$ is equal to ___

Answer 1

$\displaystyle\sum_{n=0}^{\infty} \frac{n^3((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}$
$=\displaystyle\sum_{n=0}^{\infty} \frac{1}{(n-3) !}+\displaystyle\sum_{n=0}^{\infty} \frac{3}{(n-2) !} +\sum_{n=0}^{\infty} \frac{1}{(n-1) !}+\displaystyle\sum_{n=0}^{\infty} \frac{1}{(2 n-1) !}-\displaystyle\sum_{n=0}^{\infty} \frac{1}{(2 n) !}$
$=e+3 e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$
$=5 e-\frac{1}{e}$
$\therefore a^2-b+c=26$